What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your result?
(a) `KI_(3)`
(b) `H_(2)S_(4)O_(6)`
(c)`Fe_(3)O_(4)`
(d) `CH_(3)CH_(2)OH`
(e)`CH_(3)COOH`

Explanation

(a) In `KI_3` , since the oxidation number of K is `+1` , therefore , the average oxidation number of iodine is `-1/3`.
`overset(+1)Koverset(x)I_3" "+1+3(x)=0" ":.x=-1/3`
But oxidation number cannot be fractional . In fact `KI_3` exists as : `K^(+)[I-IlarrI]^(-)` in which a coordinate bond is formed between `I_2` molecule and `I^-` . In this the oxidation number of two iodine atoms forming `I_2` molecule is zero and that of iodine forming the coordinate bond is -1. Thus, the actual oxidation number of atoms are :
`overset(+)K" "overset(0)(I_2)" "overset(-1)I`
(b) `H_2S_4O_6` . The oxidation number may be calculated as :
`overset(+1)H_2" "overset(x)(S_4)" "overset(-2)O_6`
`2(+1) +4x+6(-2)=0 " "x=+5/2`
Since oxidation number cannot be fractional , the actual oxidation number of different S atoms are :
`H-O-underset(O)underset(||)overset(O)overset(+5||" ")S-overset(0)S-overset(0)S-underset(O)underset(||)overset(O)overset(+5||" ")S-O-H`
(c) `Fe_3O_4` . The oxidation number may be calculated as :
`overset(x)Fe_3" "overset(-2)O_4`
`3x+4(-2)=0 " or " :. x = + 8/3`
In fact, `Fe_3O_4` exists as a mixture of FeO and `Fe_2O_3` as :
`overset(+2-2)(FeO).overset(+3)(Fe_2)overset(-2)O_3`
Fe has oxidation number of +2 and +3.
(d) `CH_3CH_2OH` - By conventional method , the oxidation number may be calculate as :
`overset(x)C" "overset(+1)(H_3)" "overset(x)C" "overset(+1)(H_2)" "overset(-2)O" "overset(+1)H`
`x+3(+1)+x+2(+1)+1(-2)+1(+1)=0`
`2x +4=0 " ":. x=-2`
In fact, ethanol many be written as :
`H-underset(H)underset(|)oversetHoverset(|)(C^(2))-underset(H)underset(|)oversetHoverset(|)(C^(1))-OH`
`C_2` ia attached three H atom (less eelctrogative than carbon) Therefore,
O.N. of `C_2:(3(+1)+x+1 (-1)=0 "or "x=-2`
`C_1` is attached to one OH (O.N=-1) and one `CH_3` (O.N. =+1 ) and two H = atoms
O.N. of `C_1=1+2(+1)+x+1(-1)=0 " ":. x=-2`
(e) `CH_3COOH` , By conventional method
`oversetxC" "overset(+1)H_3" "oversetxC""overset(-2)O""overset(-2)O""overset(-1)H`
` x + 3 (+1) +x + (-2) + (-2) +(1) (+1)=0`
`2x +0= 0 " ":. x = 0 `
But according to structure
`H-underset(H)underset(|)oversetHoverset(|)(C^2)-overset(O)overset(||)(C^1)-OH`
`C_2` is attached to three H atoms (less electronegative than C) and one - COOH group (more electronegative than C)
O.N. of `C_2=3(+1)+x+1(-1)=0:. x = - 2`
`C_1` is attchached to one oxygen atom by double boond , one OH - group (O.N.=-) and `CH_3` (O.N= +1)
O.N. of `C_1 = +1 +x+1+1(-2)+1(-1)=0`
`x=2=0 " or " x = pm 2`